Question

If the bisector of the angle A of triangle ABC meets Bc in D, prove that
$a=(b+c)[1-\frac{A{D}^2}{bc}{]}^{1/2}$

Answer 1

From the $△ABC$

$\frac{BD}{DC}=\frac{c}{b}=\frac{AB}{AC}$

$\Rightarrow BD=\frac{ca}{(b+c)}$

$\Rightarrow DC=\frac{b}{b+c}\times a=\frac{ab}{b+c}$

In $△ABC,\cos B=\frac{a^2+c^2-b^2}{2ac}$

In $△ABD,\cos B=\frac{c^2+{BD}^2-{AD}^2}{2c\times BD}$

$\Rightarrow \frac{a^2+c^2-b^2}{2ac}=\frac{c^2+{BD}^2-{AD}^2}{2c\times BD}$

$\Rightarrow \frac{a^2+c^2-b^2}{a}=\frac{c^2+{BD}^2-{AD}^2}{\times BD}$

$\Rightarrow \frac{a^2+c^2-b^2}{a}=\frac{c^2+{\left(\frac{ca}{(b+c)}\right)}^2-{AD}^2}{\times \left(\frac{ca}{(b+c)}\right)}$

$\Rightarrow c(a^2+c^2-b^2)=(b+c)c^2+\frac{c^2{a}^2}{b+c}-{AD}^2(b+c)$

$\Rightarrow c{a}^2+c^3-c{b}^2=c^{2}b+c^3+\frac{c^2{a}^2}{b+c}-{AD}^2(b+c)$

$\Rightarrow c{a}^2+c^3-c{b}^2-c^{2}b+c^3-\frac{c^2{a}^2}{b+c}=(b+c){AD}^2$

$\Rightarrow {AD}^2=\frac{c^2{a}^2-(b+c)(c{a}^2-c{b}^2-b{c}^2)}{{(b+c)}^2}$

$\Rightarrow {AD}^2=\frac{c^2{a}^2-bc{a}^2+b^{3}c+b^2{c}^2-c^2{a}^2+c^2{b}^2+c^{3}b}{{(b+c)}^2}$

$\Rightarrow {AD}^2=\frac{c^2{a}^2-bc{a}^2+b^{3}c+2b^2{c}^2+c^{3}b}{{(b+c)}^2}$

$\Rightarrow {AD}^2=\frac{bc[a^2-b^2-c^2-2bc]}{{(b+c)}^2}$

$\Rightarrow {AD}^2=\frac{bc[a^2-(b^2+c^2+2bc)]}{{(b+c)}^2}$

$\Rightarrow a^2-(b^2+c^2+2bc)=\frac{{AD}^2{(b+c)}^2}{bc}$

$\Rightarrow a^2=(b^2+c^2+2bc)+\frac{{AD}^2{(b+c)}^2}{bc}$

$\Rightarrow a^2={(b+c)}^2+\frac{{AD}^2{(b+c)}^2}{bc}$

$\Rightarrow a^2={(b+c)}^2[1+\frac{{AD}^2}{bc}]$

$\therefore a=\sqrt{{(b+c)}^2[1+\frac{{AD}^2}{bc}]}$

or $a=(b+c)\sqrt{[1+\frac{{AD}^2}{bc}]}$